CorollaryStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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A square root of $- 1$ in a real field extension determines a unique real-field homomorphism from $C$

Statement

Let $L / R$ be a field extension and let $j \in L$ satisfy $j^{2} = - 1$ . There is a unique field homomorphism $C \to L$ fixing $R$ and sending $i$ to $j$ . Its image is $R [j]$ .

Facts & Assumptions

Given: A real field extension $L / R$ and $j \in L$ with $j^{2} = - 1$ .

[F1]

The complex numbers are $R [x] / (x^{2} + 1)$ with $i = x + (x^{2} + 1)$ (The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ ).

[F2]

$x^{2} + 1$ is monic and irreducible over $R$ ( $x^{2} + 1$ is irreducible over $R$ ).

[F3]

In a quotient adjoining a root of a monic irreducible polynomial, any root in an extension determines a unique base-field homomorphism, whose image is the generated subring (Universal property of adjoining a root of an irreducible polynomial).

Proof

technique · direct

1.1

The equation $j^{2} = - 1$ says exactly that $j$ is a root of $x^{2} + 1$ .

algebra

2.1

Apply [F3] to [F1], [F2], and step 1.1. It gives the unique homomorphism fixing $R$ , sending $i$ to $j$ , and having image $R [j]$ .

F1F2F3step 1.1∎

Depends on

Used by

The only real-field automorphisms of ℂ are the identity and complex conjugation Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 43 results over 8 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. S. Milne, Fields and Galois Theory (standard reference, not scraped)

A square root of −1−1 in a real field extension determines a unique real-field homomorphism from CC